Question: Find the solution of the following equation whose argument is strictly between $225^\circ$ and $315^\circ$. Round your answer to the nearest thousandth. $z^7=128i$ $z$ =
Explanation: The Strategy A straightforward way to solve an equation of the form $z^{n}=m$ is by using the polar form of $z$. Therefore, our solution will consist of the following steps: Rewrite $z^n$ and $m$ in polar form. [How is this done, in general?] Solve for the modulus and argument of $z$. Find the rectangular form of $z$. [How is this done, in general?] Rewrite the equation in polar form Let's denote $r$ and $\theta$ to be the modulus and argument of $z$, respectively. Therefore, $z^{7}=r^{7}[\cos {({7}\cdot\theta)}+i\sin {({7}\cdot\theta)}]$. The number $128i$ has a modulus of $128$. The argument of $128i$ can be $90^\circ$ plus any multiple of $360^\circ$, so we can write it as $90^\circ+k\cdot360^\circ$ for an integer $k$. Now the equation looks as follows: $\begin{aligned}r^{7}[\cos {({7}\cdot\theta)}+i\sin {({7}\cdot\theta)}]&= \\128&[\cos(90^\circ+k\cdot360^\circ)+i\sin(90^\circ+k\cdot360^\circ)]\end{aligned}$ When two complex numbers are equal, we know that both their moduli and arguments are equal. Therefore, we have the following equations for $r$ and $\theta$ : $r^{7}=128$ ${7}\cdot\theta=90^\circ+k\cdot360^\circ$ Solving for $r$ $\begin{aligned}r^{7}&=128 \\\\ r &=2 \end{aligned}$ Solving for $\theta$ $\begin{aligned}{7}\cdot\theta&=90^\circ+k\cdot360^\circ \\\\\theta&=\dfrac{90}{7}^\circ+k\cdot\dfrac{360}{7}^\circ\end{aligned}$ Remember that $\theta$ is between $225^\circ$ and $315^\circ$. Therefore, we need to find the multiple of $\dfrac{360}{7}^\circ$ that is strictly within the range of $225^\circ-\dfrac{90}{7}^\circ=\dfrac{1485}{7}^\circ$ and $315^\circ-\dfrac{90}{7}^\circ=\dfrac{2115}{7}^\circ$. This multiple is simply $\dfrac{1800}{7}^\circ$, so $\theta=270^\circ$. Finding the rectangular form of $z$ Let's plug in $r=2$ and $\theta=270^\circ$ into the polar form of $z$ : $\begin{aligned}z&=r[\cos(\theta)+i\cdot\sin(\theta)]\\\\ &=2[\cos(270^\circ)+i\cdot\sin(270^\circ)]\\\\ &=2\cos(270^\circ)+2\sin(270^\circ)\cdot i\end{aligned}$ Using the calculator, we get the following solution: $z=0-2i$ Summary $z=0-2i$